1. **Problem a:** Write $v=(-5,4,9)$ as a linear combination of $u_1=(4,1,0)$ and $u_2=(1,2,3)$. 2. We want to find scalars $k_1$ and $k_2$ such that: $$v = k_1 u_1 + k_2 u_2 = k_1(4,1,0) + k_2(1,2,3) = (4k_1 + k_2, k_1 + 2k_2, 3k_2)$$ 3. Equate components: $$4k_1 + k_2 = -5$$ $$k_1 + 2k_2 = 4$$ $$3k_2 = 9$$ 4. From the third equation, solve for $k_2$: $$k_2 = \frac{9}{3} = 3$$ 5. Substitute $k_2=3$ into the first two equations: $$4k_1 + 3 = -5 \implies 4k_1 = -8 \implies k_1 = -2$$ $$k_1 + 2(3) = 4 \implies k_1 + 6 = 4 \implies k_1 = -2$$ 6. Both equations agree, so the solution is: $$k_1 = -2, \quad k_2 = 3$$ 7. Therefore, $$v = -2 u_1 + 3 u_2$$ --- **Final answer:** $v = -2(4,1,0) + 3(1,2,3)$